A ray of light is incident on a medium of refractive index ` sqrt(2)` at an angle of incidence of `45^(@)`. The ratio of the width of the incident beam in air to that of the refracted beam in the medium is

Options

A. `(3//2)^(1//2)`
B. `(2//3)^(1//2)`
C. `3//2`
D. `2//3`

Explanation

To solve the problem, we need to find the ratio of the width of the incident beam in air to that of the refracted beam in the medium with a refractive index of \( \sqrt{2} \) when the angle of incidence is \( 45^\circ \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Refractive index of the medium, \( n = \sqrt{2} \)
- Angle of incidence, \( i = 45^\circ \)

2. **Use Snell's Law:**
Snell's Law states that:
\[ n_1 \sin(i) = n_2 \sin(r) \]
Here, \( n_1 = 1 \) (for air), \( n_2 = \sqrt{2} \), and \( r \) is the angle of refraction. Plugging in the values:
\[ 1 \cdot \sin(45^\circ) = \sqrt{2} \cdot \sin(r) \]
Since \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ \frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin(r) \]
Rearranging gives:
\[ \sin(r) = \frac{1}{2} \]
Thus, \( r = 30^\circ \).

3. **Calculate the Widths:**
The width of the incident beam in air is denoted as \( AB \) and the width of the refracted beam in the medium is denoted as \( CD \). The ratio of the widths can be found using the cosine of the angles:
\[ \frac{AB}{CD} = \frac{\cos(i)}{\cos(r)} \]

4. **Calculate \( \cos(i) \) and \( \cos(r) \):**
- \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)

5. **Substitute the Values:**
Now substituting these values into the ratio:
\[ \frac{AB}{CD} = \frac{\cos(45^\circ)}{\cos(30^\circ)} = \frac{\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{2}} \cdot \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{6}} = \frac{2}{\sqrt{2 \cdot 3}} = \frac{2}{\sqrt{6}} = \frac{2 \sqrt{6}}{6} = \frac{\sqrt{6}}{3} \]

6. **Final Ratio:**
To express this in a more standard form, we can rationalize the denominator:
\[ \frac{AB}{CD} = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3} \]
However, we can also express this as:
\[ \frac{AB}{CD} = \frac{2}{3^{1/2}} = \frac{2}{3^{1/2}} = \frac{2}{3^{1/2}} \text{ (which is equivalent to } \frac{2}{3^{1/2}} \text{)} \]

Thus, the final answer for the ratio of the width of the incident beam in air to that of the refracted beam in the medium is:
\[ \frac{2}{3^{1/2}} \]